Answer
(a) $74.6Hz$
(b) $4.61W$
Work Step by Step
(a) As $Z=\frac{R}{cos\phi}=\frac{3.35K\Omega}{cos23.0^{\circ}}=3.639K\Omega$
We know that
$f=\frac{1}{2\pi C\sqrt{Z^2-R^2}}$
We plug in the known values to obtain:
$f=\frac{1}{2\pi(1.50\times 10^{-6})\sqrt{(3.639K\Omega)^2-(3.35K\Omega)^2}}=74.6Hz$
(b) We can find the required power as
$P_{avg}=I_{rms}^2 R$
We plug in the known values to obtain:
$P_{avg}=(\frac{V_{rms}}{Z})^2 R$
We plug in the known values to obtain:
$P_{avg}=(\frac{135}{3.63K\Omega})^2(3.35K\Omega)^2=4.61W$